# $\int_0^\infty f(x) \; dx < \infty$ implies $\lim_{x \rightarrow \infty} x f(x) = 0$. [duplicate]

Let $f$ be non-negative, monotone decreasing such that $$\int_0^\infty f(x) \; dx < \infty$$ Show that $$\lim_{x \rightarrow \infty} x f(x) = 0.$$

I have the following solution, but wonder if there is a better way.

Let $\epsilon > 0$ be small. Suppose to the contrary that $\limsup x f(x) = c > 0$, then there is a sequence $x_n$ tending to infinity such that $x_n f(x_n) \geq c - \epsilon$. Moreover, $f(x) \geq (c-\epsilon)/x_n$ for $x \leq x_n$. Thus, the integral $\int_0^N f(x) \; dx$ is bounded from below by $$S_N = x_1 \frac{c-\epsilon}{x_1} + (x_2 - x_1) \frac{c-\epsilon}{x_2} + \cdots + (x_N - x_{N-1}) \frac{c-\epsilon}{x_N}$$ Since $x_n \rightarrow \infty$, we may suppose that the $x_n$ are spaced far enough apart so that $x_k/x_{k+1} < M < 1$. Then the partial sum $S_N$ will be bounded below by $$N(c-\epsilon) - (N-1)M(c-\epsilon)$$ which tends to infinity as $N \rightarrow \infty$.

Is there an easier solution?

## marked as duplicate by 5xum, Andrés E. Caicedo, Daniel Fischer, drhab, Jonas MeyerJul 22 '14 at 18:44

• I feel like the referenced answer might be a bit overkill. I might be being stupid, but can't we just do some kind of comparison to $1/x$ ? – JC574 Jul 22 '14 at 18:39
• Limit comparison requires $f$ to be continuous, and for the limit $f(x)/x$ to exist, which it may not. You can try adapting the proof, which is more or less what I tried to do. – user09812093 Jul 22 '14 at 18:42
• @JC574 I think that would work, because the convergence of the intergal implies that $\exists x_0: f<\frac{1}{x}\; \forall (x>x_0>0)\implies x<\frac{1}{f(x)} \forall x>x_0 \implies x=o(f)$ – user76844 Jul 22 '14 at 18:44
• @user09812093 I don't think that's what i meant - $f$ is monotonic decreasing. If there existed $\epsilon > 0$ such that $f$ was always greater than $\epsilon/x$ our integral wouldn't converge would it? or is that wrong? – JC574 Jul 22 '14 at 18:48